A2 Review Unit 4 Quadratic functions. Choose a topic.
Graph of Quadratic Functions Questions 1 – 4. Choose a question.
Transformations of Quadratic Functions Questions 5 – 7. Choose a question.
List the transformations, give the vertex, and rewrite the equation in standard form.
y = (x − 2)² + 6
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Question 5 · Transformations y = (x − 2)² + 6
Shift right 2 units, shift up 6 units
a = 1.Since 1 is positive there is no reflection over the x -axis. Since |1| = 1 there is no stretch and no compression. h = 2. There is a horizontal shift to the right 2 units.k = 6. There is a vertical shift up 6 units.Vertex Standard form Question 5 · Vertex y = (x − 2)² + 6
Vertex: (2, 6)
In vertex form y = a (x − h )² + k , the vertex is (h , k ). h = 2 and k = 6, so the vertex is (2, 6).Transformations Standard form Question 5 · Standard Form y = (x − 2)² + 6
y = x ² − 4x + 10
y = (x − 2)(x − 2) + 6y = (x ² − 2x − 2x + 4) + 6y = x ² − 4x + 4 + 6y = x ² − 4x + 10Transformations Vertex List the transformations, give the vertex, and rewrite the equation in standard form.
y = −3(x + 5)² − 7
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Question 6 · Transformations y = −3(x + 5)² − 7
Reflection, stretch by 3, left 5, down 7
a = −3.Since −3 is negative there is a reflection over the x -axis. Since |−3| > 1 there is a stretch by a factor of 3. h = −5. There is a horizontal shift to the left 5 units.k = −7. There is a vertical shift down 7 units.Vertex Standard form Question 6 · Vertex y = −3(x + 5)² − 7
Vertex: (−5, −7)
In vertex form y = a (x − h )² + k , the vertex is (h , k ). h = −5 and k = −7, so the vertex is (−5, −7).Transformations Standard form Question 6 · Standard Form y = −3(x + 5)² − 7
y = −3x ² − 30x − 82
y = −3(x + 5)(x + 5) − 7y = −3(x ² + 5x + 5x + 25) − 7y = −3x ² − 15x − 15x − 75 − 7y = −3x ² − 30x − 82Transformations Vertex List the transformations, give the vertex, and rewrite the equation in standard form.
y = 1 2 x − 4)²
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Question 7 · Transformations y = 1 2 x − 4)²
Compression by 1 2
a = 1 2 Since 1 2 x -axis. Since |1 2 1 2 h = 4. There is a horizontal shift to the right 4 units.k = 0. There is no vertical shift.Vertex Standard form Question 7 · Vertex y = 1 2 x − 4)²
Vertex: (4, 0)
In vertex form y = a (x − h )² + k , the vertex is (h , k ). y = 1 2 x − 4)² + 0, so h = 4 and k = 0. The vertex is (4, 0).Transformations Standard form Question 7 · Standard Form y = 1 2 x − 4)²
y = 1 2 x ² − 4x + 8
y = 1 2 x − 4)(x − 4)y = 1 2 x ² − 4x − 4x + 16)y = 1 2 x ² − 2x − 2x + 8y = 1 2 x ² − 4x + 8Transformations Vertex Solve Quadratic Equations by Graphing Questions 8 – 10. Choose a question.
Solve the quadratic equation by graphing.
2(x + 2)² − 3 = 0
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Question 8 · Graph & Solutions 2(x + 2)² − 3 = 0
x ≈ −0.5 and x ≈ −3.5
Vertex form: the vertex is (−2, −3); a = 2 > 0, so the graph opens up. Choose x = −1: y = 2(1)² − 3 = −1 → (−1, −1); reflect: (−3, −1). Choose x = 0: y = 2(2)² − 3 = 5 → (0, 5); reflect: (−4, 5). The solutions of the equation are the x -intercepts. The graph crosses the x -axis between 0 and −1 and between −3 and −4. The solutions are approximately −0.5 and −3.5. Solve the quadratic equation by graphing.
x ² − 8x + 18 = 0
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Question 9 · Graph & Solutions x ² − 8x + 18 = 0
No real solution
Vertex: x = −b /(2a ) = −(−8)/(2·1) = 4; y = 4² − 8(4) + 18 = 2. Vertex (4, 2). a = 1 > 0, so the graph opens up.Choose x = 3: y = 3 → (3, 3); reflect: (5, 3). Choose x = 2: y = 6 → (2, 6); reflect: (6, 6). The solutions of the equation are the x -intercepts. The vertex (4, 2) is above the x -axis and the graph opens up, so the graph never crosses the x -axis. There is no real solution. Solve the quadratic equation by graphing.
−3(x − 4)² = 0
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Question 10 · Graph & Solutions −3(x − 4)² = 0
x = 4 (one solution)
Vertex form: the vertex is (4, 0); a = −3 < 0, so the graph opens down. |−3| > 1, so the graph is narrower than y = x ². Choose x = 3: y = −3(−1)² = −3 → (3, −3); reflect: (5, −3). The solutions of the equation are the x -intercepts. The graph touches the x -axis only at its vertex (4, 0). There is one solution: x = 4. Factor Completely Quadratic Expressions Questions 11 – 20. Choose a question.
Question 11 4 steps
Factor completely Try the first step on paper, then press Next to check it.
x ² + 21x + 20
Question 11 · First, check for a GCF Step 1 of 4
First, check for a GCF The coefficient of x ² is 1, so the only common number factor is 1, and the last term 20 has no x . No GCF to factor out .
GCF = 1
Original expression: x ² + 21x + 20
Question 11 · Identify b and c Step 2 of 4
Identify b and c The coefficient of x ² is 1 , so we can use the shortcut : find two numbers r and s whose product is c and whose sum is b . Then x ² + b x + c factors directly as (x + r )(x + s ) . Read b and c , keeping their signs.
a = 1, b = 21, c = 20
Original expression: x ² + 21x + 20
Question 11 · Find the pair Step 3 of 4
Find the two numbers Find two numbers that multiply to 20 and add to 21 . The product is positive, so both numbers have the same sign . The sum is positive, so both numbers are positive. Factor pairs of 20: 1 · 20, 2 · 10, 4 · 5. The pair 1 and 20 works.
(1 )(20 ) = 20 1 + 20 = 21
Original expression: x ² + 21x + 20
Question 11 · Write the factors Step 4 of 4
Write the factors Put the two numbers, 1 and 20, straight into the pattern (x + ▢)(x + ▢).
(x + 1 )(x + 20 )
Original expression: x ² + 21x + 20
Question 11 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
x ² + 21x + 20 = (x + 1)(x + 20)
Original expression: x ² + 21x + 20
Question 12 4 steps
Factor completely Try the first step on paper, then press Next to check it.
x ² − 9x + 20
Question 12 · First, check for a GCF Step 1 of 4
First, check for a GCF The coefficient of x ² is 1, so the only common number factor is 1, and the last term 20 has no x . No GCF to factor out .
GCF = 1
Original expression: x ² − 9x + 20
Question 12 · Identify b and c Step 2 of 4
Identify b and c The coefficient of x ² is 1 , so we can use the shortcut : find two numbers r and s whose product is c and whose sum is b . Then x ² + b x + c factors directly as (x + r )(x + s ) . Read b and c , keeping their signs.
a = 1, b = −9, c = 20
Original expression: x ² − 9x + 20
Question 12 · Find the pair Step 3 of 4
Find the two numbers Find two numbers that multiply to 20 and add to −9 . The product is positive, so both numbers have the same sign . The sum is negative, so both numbers are negative. Factor pairs of 20: 1 · 20, 2 · 10, 4 · 5. The pair −4 and −5 works.
(−4 )(−5 ) = 20 −4 + (−5 ) = −9
Original expression: x ² − 9x + 20
Question 12 · Write the factors Step 4 of 4
Write the factors Put the two numbers, −4 and −5, straight into the pattern (x + ▢)(x + ▢). Adding a negative number is the same as subtracting.
(x + (−4 ))(x + (−5 )) → (x − 4)(x − 5)
Original expression: x ² − 9x + 20
Question 12 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
x ² − 9x + 20 = (x − 4)(x − 5)
Original expression: x ² − 9x + 20
Question 13 7 steps
Factor completely Try the first step on paper, then press Next to check it.
2x ² + 5x − 3
Question 13 · First, check for a GCF Step 1 of 7
First, check for a GCF List the coefficients 2, 5 and −3. Their only common factor is 1, and not every term has the variable. No GCF to factor out . The expression is a quadratic trinomial (an expression of the form a x ² + b x + c ), so it needs to be factored using the AC method .
GCF(2, 5, 3) = 1
Original expression: 2x ² + 5x − 3
Question 13 · Multiply a ·c Step 2 of 7
Identify a , b and c , then multiply a · c Read the coefficients, then multiply the first and last. Keep the sign of c .
a = 2, b = 5, c = −3 a · c = 2 · (−3) = −6
Original expression: 2x ² + 5x − 3
Question 13 · Find the pair Step 3 of 7
Find two numbers Find two numbers that multiply to −6 and add to 5 . The product is negative, so one number is positive and one is negative . The sum is positive, so the bigger number (ignoring signs) is the positive one. Factor pairs of 6: 1 · 6, 2 · 3. The pair 6 and −1 works.
(6 )(−1 ) = −6 6 + (−1 ) = 5
Original expression: 2x ² + 5x − 3
Question 13 · Split the middle Step 4 of 7
Split the middle term Rewrite 5x as 6x − x .
2x ² + 6x − x − 3
Original expression: 2x ² + 5x − 3
Question 13 · Factor by grouping Step 5 of 7
Factor the first two terms and the last two terms From the first two terms, 2x ² + 6x , factor out 2x . From the last two terms, −x − 3, factor out −1 . We factor out −1 (with its negative sign) so the second parentheses match the first.
2x (x + 3) − 1(x + 3)
Original expression: 2x ² + 5x − 3
Question 13 · Factor by grouping Step 6 of 7
Spot the common factor Look inside the parentheses. Both terms have the same binomial, (x + 3) . Because what is inside the parentheses is the same in both terms, (x + 3) is a common factor of the whole expression.
2x (x + 3x + 3
Original expression: 2x ² + 5x − 3
Question 13 · Factor by grouping Step 7 of 7
Factor out the common binomial Factor (x + 3) out of both terms.
(x + 3)(2x − 1)
Original expression: 2x ² + 5x − 3
Question 13 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
2x ² + 5x − 3 = (x + 3)(2x − 1)
Original expression: 2x ² + 5x − 3
Question 14 7 steps
Factor completely Try the first step on paper, then press Next to check it.
4x ² − 3x − 1
Question 14 · First, check for a GCF Step 1 of 7
First, check for a GCF List the coefficients 4, −3 and −1. Their only common factor is 1, and not every term has the variable. No GCF to factor out . The expression is a quadratic trinomial (an expression of the form a x ² + b x + c ), so it needs to be factored using the AC method .
GCF(4, 3, 1) = 1
Original expression: 4x ² − 3x − 1
Question 14 · Multiply a ·c Step 2 of 7
Identify a , b and c , then multiply a · c Read the coefficients, then multiply the first and last. Keep the sign of c .
a = 4, b = −3, c = −1 a · c = 4 · (−1) = −4
Original expression: 4x ² − 3x − 1
Question 14 · Find the pair Step 3 of 7
Find two numbers Find two numbers that multiply to −4 and add to −3 . The product is negative, so one number is positive and one is negative . The sum is negative, so the bigger number (ignoring signs) is the negative one. Factor pairs of 4: 1 · 4, 2 · 2. The pair −4 and 1 works.
(−4 )(1 ) = −4 −4 + 1 = −3
Original expression: 4x ² − 3x − 1
Question 14 · Split the middle Step 4 of 7
Split the middle term Rewrite −3x as −4x + x .
4x ² − 4x + x − 1
Original expression: 4x ² − 3x − 1
Question 14 · Factor by grouping Step 5 of 7
Factor the first two terms and the last two terms From the first two terms, 4x ² − 4x , factor out 4x . The last two terms, x − 1, have no common factor except 1 , so factor out 1. That keeps the parentheses matching.
4x (x − 1) + 1(x − 1)
Original expression: 4x ² − 3x − 1
Question 14 · Factor by grouping Step 6 of 7
Spot the common factor Look inside the parentheses. Both terms have the same binomial, (x − 1) . Because what is inside the parentheses is the same in both terms, (x − 1) is a common factor of the whole expression.
4x (x − 1x − 1
Original expression: 4x ² − 3x − 1
Question 14 · Factor by grouping Step 7 of 7
Factor out the common binomial Factor (x − 1) out of both terms.
(x − 1)(4x + 1)
Original expression: 4x ² − 3x − 1
Question 14 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
4x ² − 3x − 1 = (x − 1)(4x + 1)
Original expression: 4x ² − 3x − 1
Question 15 3 steps
Factor completely Try the first step on paper, then press Next to check it.
x ² − 169
Question 15 · First, check for a GCF Step 1 of 3
First, check for a GCF The coefficients 1 and −169 share only 1, and the constant has no x . No GCF to factor out .
GCF(1, 169) = 1
Original expression: x ² − 169
Question 15 · Recognize the pattern Step 2 of 3
Recognize the pattern The expression has two terms , both are perfect squares , and they are subtracted . It is a difference of squares : a ² − b ² = (a − b )(a + b ).
x ² − 169 = (x 13 )²
Original expression: x ² − 169
Question 15 · Write the factors Step 3 of 3
Write the factors Here a = x and b = 13. Write (a − b )(a + b ).
(x − 13)(x + 13)
Original expression: x ² − 169
Question 15 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
x ² − 169 = (x − 13)(x + 13)
Original expression: x ² − 169
Question 16 3 steps
Factor completely Try the first step on paper, then press Next to check it.
25x ² − 16
Question 16 · First, check for a GCF Step 1 of 3
First, check for a GCF The coefficients 25 and −16 share only 1, and the constant has no x . No GCF to factor out .
GCF(25, 16) = 1
Original expression: 25x ² − 16
Question 16 · Recognize the pattern Step 2 of 3
Recognize the pattern The expression has two terms , both are perfect squares , and they are subtracted . It is a difference of squares : a ² − b ² = (a − b )(a + b ). 25x ² = (5x )² and 16 = 4².
25x ² − 16 = (5x )² − (4 )²
Original expression: 25x ² − 16
Question 16 · Write the factors Step 3 of 3
Write the factors Here a = 5x and b = 4. Write (a − b )(a + b ).
(5x − 4)(5x + 4)
Original expression: 25x ² − 16
Question 16 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
25x ² − 16 = (5x − 4)(5x + 4)
Original expression: 25x ² − 16
Question 17 9 steps
Factor completely Try the first step on paper, then press Next to check it.
20x ² − 54x + 36
Question 17 · First, check for a GCF Step 1 of 9
First, check for a GCF Coefficients 20, −54 and 36 are all divisible by 2 . Not every term has x . So the GCF is 2.
GCF(20, 54, 36) = 2
Original expression: 20x ² − 54x + 36
Question 17 · First, check for a GCF Step 2 of 9
Factor out the GCF Divide every term by 2: 20x ² ÷ 2 = 10x ², −54x ÷ 2 = −27x , 36 ÷ 2 = 18.
2 (10x ² − 27x + 18)
Original expression: 20x ² − 54x + 36
Question 17 · First, check for a GCF Step 3 of 9
Check what's left What is left inside the parentheses is a quadratic trinomial (an expression of the form a x ² + b x + c ). It needs to be factored using the AC method .
2(10x ² − 27x + 18 )
Original expression: 20x ² − 54x + 36
Question 17 · Multiply a ·c Step 4 of 9
Identify a , b and c , then multiply a · c Read a , b and c from the trinomial inside the parentheses.
a = 10, b = −27, c = 18 a · c = 10 · 18 = 180
Original expression: 20x ² − 54x + 36
Question 17 · Find the pair Step 5 of 9
Find two numbers Find two numbers that multiply to 180 and add to −27 . The product is positive, so both numbers have the same sign . The sum is negative, so both numbers are negative. Factor pairs of 180: 1 · 180, 2 · 90, 3 · 60, 4 · 45, 5 · 36, 6 · 30, 9 · 20, 10 · 18, 12 · 15. The pair −12 and −15 works.
(−12 )(−15 ) = 180 −12 + (−15 ) = −27
Original expression: 20x ² − 54x + 36
Question 17 · Split the middle Step 6 of 9
Split the middle term Rewrite −27x as −12x − 15x .
2(10x ² − 12x − 15x + 18)
Original expression: 20x ² − 54x + 36
Question 17 · Factor by grouping Step 7 of 9
Factor the first two terms and the last two terms From the first two terms, 10x ² − 12x , factor out 2x . From the last two terms, −15x + 18, factor out −3 . We factor out −3 (with its negative sign) so the second parentheses match the first.
2[2x (5x − 6) − 3(5x − 6)]
Original expression: 20x ² − 54x + 36
Question 17 · Factor by grouping Step 8 of 9
Spot the common factor Look inside the parentheses. Both terms have the same binomial, (5x − 6) . Because what is inside the parentheses is the same in both terms, (5x − 6) is a common factor of the whole expression.
2[2x (5x − 6 ) − 3(5x − 6 )]
Original expression: 20x ² − 54x + 36
Question 17 · Factor by grouping Step 9 of 9
Factor out the common binomial Factor (5x − 6) out of both terms. Keep the 2 in front.
2(5x − 6)(2x − 3)
Original expression: 20x ² − 54x + 36
Question 17 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
20x ² − 54x + 36 = 2(5x − 6)(2x − 3)
Original expression: 20x ² − 54x + 36
Question 18 9 steps
Factor completely Try the first step on paper, then press Next to check it.
42x ² + 35x + 7
Question 18 · First, check for a GCF Step 1 of 9
First, check for a GCF Coefficients 42, 35 and 7 are all divisible by 7 . Not every term has x . So the GCF is 7.
GCF(42, 35, 7) = 7
Original expression: 42x ² + 35x + 7
Question 18 · First, check for a GCF Step 2 of 9
Factor out the GCF Divide every term by 7: 42x ² ÷ 7 = 6x ², 35x ÷ 7 = 5x , 7 ÷ 7 = 1. Don't forget the 1 at the end.
7 (6x ² + 5x + 1)
Original expression: 42x ² + 35x + 7
Question 18 · First, check for a GCF Step 3 of 9
Check what's left What is left inside the parentheses is a quadratic trinomial (an expression of the form a x ² + b x + c ). It needs to be factored using the AC method .
7(6x ² + 5x + 1 )
Original expression: 42x ² + 35x + 7
Question 18 · Multiply a ·c Step 4 of 9
Identify a , b and c , then multiply a · c Read a , b and c from the trinomial inside the parentheses.
a = 6, b = 5, c = 1 a · c = 6 · 1 = 6
Original expression: 42x ² + 35x + 7
Question 18 · Find the pair Step 5 of 9
Find two numbers Find two numbers that multiply to 6 and add to 5 . The product is positive, so both numbers have the same sign . The sum is positive, so both numbers are positive. Factor pairs of 6: 1 · 6, 2 · 3. The pair 2 and 3 works.
(2 )(3 ) = 6 2 + 3 = 5
Original expression: 42x ² + 35x + 7
Question 18 · Split the middle Step 6 of 9
Split the middle term Rewrite 5x as 2x + 3x .
7(6x ² + 2x + 3x + 1)
Original expression: 42x ² + 35x + 7
Question 18 · Factor by grouping Step 7 of 9
Factor the first two terms and the last two terms From the first two terms, 6x ² + 2x , factor out 2x . From the last two terms, 3x + 1, the only common factor is 1 , so factor out 1.
7[2x (3x + 1) + 1(3x + 1)]
Original expression: 42x ² + 35x + 7
Question 18 · Factor by grouping Step 8 of 9
Spot the common factor Look inside the parentheses. Both terms have the same binomial, (3x + 1) . Because what is inside the parentheses is the same in both terms, (3x + 1) is a common factor of the whole expression.
7[2x (3x + 1 ) + 1(3x + 1 )]
Original expression: 42x ² + 35x + 7
Question 18 · Factor by grouping Step 9 of 9
Factor out the common binomial Factor (3x + 1) out of both terms. Keep the 7 in front.
7(3x + 1)(2x + 1)
Original expression: 42x ² + 35x + 7
Question 18 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
42x ² + 35x + 7 = 7(3x + 1)(2x + 1)
Original expression: 42x ² + 35x + 7
Question 19 4 steps
Factor completely Try the first step on paper, then press Next to check it.
27x ² − 3
Question 19 · First, check for a GCF Step 1 of 4
First, check for a GCF Coefficients 27 and −3 are both divisible by 3 . The constant has no x . So the GCF is 3.
GCF(27, 3) = 3
Original expression: 27x ² − 3
Question 19 · First, check for a GCF Step 2 of 4
Factor out the GCF Divide every term by 3: 27x ² ÷ 3 = 9x ², −3 ÷ 3 = −1.
3 (9x ² − 1)
Original expression: 27x ² − 3
Question 19 · Recognize the pattern Step 3 of 4
Recognize the pattern The expression has two terms , both are perfect squares , and they are subtracted . It is a difference of squares : a ² − b ² = (a − b )(a + b ). 9x ² = (3x )² and 1 = 1².
3[(3x )² − (1 )²]
Original expression: 27x ² − 3
Question 19 · Write the factors Step 4 of 4
Write the factors Here a = 3x and b = 1. Write (a − b )(a + b ). Keep the 3 in front.
3(3x − 1)(3x + 1)
Original expression: 27x ² − 3
Question 19 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
27x ² − 3 = 3(3x − 1)(3x + 1)
Original expression: 27x ² − 3
Question 20 4 steps
Factor completely Try the first step on paper, then press Next to check it.
18x ² − 8
Question 20 · First, check for a GCF Step 1 of 4
First, check for a GCF Coefficients 18 and −8 are both divisible by 2 . The constant has no x . So the GCF is 2.
GCF(18, 8) = 2
Original expression: 18x ² − 8
Question 20 · First, check for a GCF Step 2 of 4
Factor out the GCF Divide every term by 2: 18x ² ÷ 2 = 9x ², −8 ÷ 2 = −4.
2 (9x ² − 4)
Original expression: 18x ² − 8
Question 20 · Recognize the pattern Step 3 of 4
Recognize the pattern The expression has two terms , both are perfect squares , and they are subtracted . It is a difference of squares : a ² − b ² = (a − b )(a + b ). 9x ² = (3x )² and 4 = 2².
2[(3x )² − (2 )²]
Original expression: 18x ² − 8
Question 20 · Write the factors Step 4 of 4
Write the factors Here a = 3x and b = 2. Write (a − b )(a + b ). Keep the 2 in front.
2(3x − 2)(3x + 2)
Original expression: 18x ² − 8
Question 20 · Done Done
Factored completely Nice work. Press Back to Questions to choose another question.
18x ² − 8 = 2(3x − 2)(3x + 2)
Original expression: 18x ² − 8
Solve Quadratic Equations by Factoring Questions 21 – 25. Choose a question.
Question 21 4 steps
Solve by factoring Try the first step on paper, then press Next to check it.
6x ² + 5x = 0
Question 21 · Standard form Step 1 of 4
Check standard form 0 is already alone on the right. There is no constant term, so c = 0. Do not divide both sides by x ; that would lose the solution x = 0.
6x ² + 5x = 0
Original equation: 6x ² + 5x = 0
Question 21 · First, check for a GCF Step 2 of 4
First, check for a GCF Coefficients 6 and 5 share only 1, but both terms have an x x .
GCF(6x ², 5x ) = x
Original equation: 6x ² + 5x = 0
Question 21 · First, check for a GCF Step 3 of 4
Factor out the GCF Divide each term by x : 6x ² ÷ x = 6x , and 5x ÷ x = 5. What's left inside the parentheses is a linear binomial (x is not raised to any power greater than 1). It cannot be factored further, so you are done.
x x + 5) = 0
Original equation: 6x ² + 5x = 0
Question 21 · Solve Step 4 of 4
Set each factor equal to zero Zero product property: if a product is 0, one of the factors is 0. Solve each small equation.
x = 0 or 6x + 5 = 0
Original equation: 6x ² + 5x = 0
Question 21 · Done Done
Solution Nice work. Press Back to Questions to choose another question.
x = 0 or x = [−5/6]
Original equation: 6x ² + 5x = 0
Question 22 9 steps
Solve by factoring Try the first step on paper, then press Next to check it.
2x ² − 3x − 9 = 0
Question 22 · Standard form Step 1 of 9
Check standard form The equation is already in the form a x ² + b x + c = 0.
2x ² − 3x − 9 = 0
Original equation: 2x ² − 3x − 9 = 0
Question 22 · First, check for a GCF Step 2 of 9
First, check for a GCF List the coefficients 2, −3 and −9. Their only common factor is 1, and not every term has the variable. No GCF to factor out . The expression is a quadratic trinomial (an expression of the form a x ² + b x + c ), so it needs to be factored using the AC method .
GCF(2, 3, 9) = 1
Original equation: 2x ² − 3x − 9 = 0
Question 22 · Multiply a ·c Step 3 of 9
Identify a , b and c , then multiply a · c Read the coefficients, then multiply the first and last. Keep the sign of c .
a = 2, b = −3, c = −9 a · c = 2 · (−9) = −18
Original equation: 2x ² − 3x − 9 = 0
Question 22 · Find the pair Step 4 of 9
Find two numbers Find two numbers that multiply to −18 and add to −3 . The product is negative, so one number is positive and one is negative . The sum is negative, so the bigger number (ignoring signs) is the negative one. Factor pairs of 18: 1 · 18, 2 · 9, 3 · 6. The pair −6 and 3 works.
(−6 )(3 ) = −18 −6 + 3 = −3
Original equation: 2x ² − 3x − 9 = 0
Question 22 · Split the middle Step 5 of 9
Split the middle term Rewrite −3x as −6x + 3x .
2x ² − 6x + 3x − 9 = 0
Original equation: 2x ² − 3x − 9 = 0
Question 22 · Factor by grouping Step 6 of 9
Factor the first two terms and the last two terms From the first two terms, 2x ² − 6x , factor out 2x . From the last two terms, 3x − 9, factor out 3 .
2x (x − 3) + 3(x − 3) = 0
Original equation: 2x ² − 3x − 9 = 0
Question 22 · Factor by grouping Step 7 of 9
Spot the common factor Look inside the parentheses. Both terms have the same binomial, (x − 3) . Because what is inside the parentheses is the same in both terms, (x − 3) is a common factor of the whole expression.
2x (x − 3x − 3
Original equation: 2x ² − 3x − 9 = 0
Question 22 · Factor by grouping Step 8 of 9
Factor out the common binomial Factor (x − 3) out of both terms.
(x − 3)(2x + 3) = 0
Original equation: 2x ² − 3x − 9 = 0
Question 22 · Solve Step 9 of 9
Set each factor equal to zero Zero product property: if a product is 0, one of the factors is 0. Solve each small equation.
x − 3 = 0 or 2x + 3 = 0
Original equation: 2x ² − 3x − 9 = 0
Question 22 · Done Done
Solution Nice work. Press Back to Questions to choose another question.
x = 3 or x = −3 2
Original equation: 2x ² − 3x − 9 = 0
Question 23 10 steps
Solve by factoring Try the first step on paper, then press Next to check it.
18x ² + 48x = −32
Question 23 · Standard form Step 1 of 10
Write in standard form The right side must be 0 before you factor. Add 32 to both sides.
18x ² + 48x + 32 = 0
Original equation: 18x ² + 48x = −32
Question 23 · First, check for a GCF Step 2 of 10
First, check for a GCF Coefficients 18, 48 and 32 are all divisible by 2 . Factor it out: 18x ² → 9x ², 48x → 24x , 32 → 16.
2 (9x ² + 24x + 16) = 0
Original equation: 18x ² + 48x = −32
Question 23 · First, check for a GCF Step 3 of 10
Check what's left What is left inside the parentheses is a quadratic trinomial (an expression of the form a x ² + b x + c ). It needs to be factored using the AC method .
2(9x ² + 24x + 16 ) = 0
Original equation: 18x ² + 48x = −32
Question 23 · Multiply a ·c Step 4 of 10
Identify a , b and c , then multiply a · c Read a , b and c from the trinomial inside the parentheses.
a = 9, b = 24, c = 16 a · c = 9 · 16 = 144
Original equation: 18x ² + 48x = −32
Question 23 · Find the pair Step 5 of 10
Find two numbers Find two numbers that multiply to 144 and add to 24 . The product is positive, so both numbers have the same sign . The sum is positive, so both numbers are positive. Factor pairs of 144: 1 · 144, 2 · 72, 3 · 48, 4 · 36, 6 · 24, 8 · 18, 9 · 16, 12 · 12. The pair 12 and 12 works.
(12 )(12 ) = 144 12 + 12 = 24
Original equation: 18x ² + 48x = −32
Question 23 · Split the middle Step 6 of 10
Split the middle term Rewrite 24x as 12x + 12x .
2(9x ² + 12x + 12x + 16) = 0
Original equation: 18x ² + 48x = −32
Question 23 · Factor by grouping Step 7 of 10
Factor the first two terms and the last two terms From the first two terms, 9x ² + 12x , factor out 3x . From the last two terms, 12x + 16, factor out 4 .
2[3x (3x + 4) + 4(3x + 4)] = 0
Original equation: 18x ² + 48x = −32
Question 23 · Factor by grouping Step 8 of 10
Spot the common factor Look inside the parentheses. Both terms have the same binomial, (3x + 4) . Because what is inside the parentheses is the same in both terms, (3x + 4) is a common factor of the whole expression.
2[3x (3x + 4 ) + 4(3x + 4 )] = 0
Original equation: 18x ² + 48x = −32
Question 23 · Factor by grouping Step 9 of 10
Factor out the common binomial Factor (3x + 4) out. Keep the 2 in front. The same factor appears twice, so it can be written as a square.
2(3x + 4)(3x + 4) = 0 → 2(3x + 4)² = 0
Original equation: 18x ² + 48x = −32
Question 23 · Solve Step 10 of 10
Set each factor equal to zero The factor 2 can never be 0, so it gives no solution. Both other factors are 3x + 4, so there is only one solution .
3x + 4 = 0
Original equation: 18x ² + 48x = −32
Question 23 · Done Done
Solution Nice work. Press Back to Questions to choose another question.
x = −4 3
Original equation: 18x ² + 48x = −32
Question 24 10 steps
Solve by factoring Try the first step on paper, then press Next to check it.
5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Standard form Step 1 of 10
Write in standard form The right side must be 0 before you factor. Add 3x ² to both sides and subtract 6x from both sides. Combine like terms.
8x ² − 8x − 6 = 0
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · First, check for a GCF Step 2 of 10
First, check for a GCF Coefficients 8, −8 and −6 are all divisible by 2 . Factor it out: 8x ² → 4x ², −8x → −4x , −6 → −3.
2 (4x ² − 4x − 3) = 0
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · First, check for a GCF Step 3 of 10
Check what's left What is left inside the parentheses is a quadratic trinomial (an expression of the form a x ² + b x + c ). It needs to be factored using the AC method .
2(4x ² − 4x − 3 ) = 0
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Multiply a ·c Step 4 of 10
Identify a , b and c , then multiply a · c Read a , b and c from the trinomial inside the parentheses.
a = 4, b = −4, c = −3 a · c = 4 · (−3) = −12
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Find the pair Step 5 of 10
Find two numbers Find two numbers that multiply to −12 and add to −4 . The product is negative, so one number is positive and one is negative . The sum is negative, so the bigger number (ignoring signs) is the negative one. Factor pairs of 12: 1 · 12, 2 · 6, 3 · 4. The pair −6 and 2 works.
(−6 )(2 ) = −12 −6 + 2 = −4
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Split the middle Step 6 of 10
Split the middle term Rewrite −4x as −6x + 2x .
2(4x ² − 6x + 2x − 3) = 0
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Factor by grouping Step 7 of 10
Factor the first two terms and the last two terms From the first two terms, 4x ² − 6x , factor out 2x . The last two terms, 2x − 3, have no common factor except 1 , so factor out 1.
2[2x (2x − 3) + 1(2x − 3)] = 0
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Factor by grouping Step 8 of 10
Spot the common factor Look inside the parentheses. Both terms have the same binomial, (2x − 3) . Because what is inside the parentheses is the same in both terms, (2x − 3) is a common factor of the whole expression.
2[2x (2x − 3 ) + 1(2x − 3 )] = 0
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Factor by grouping Step 9 of 10
Factor out the common binomial Factor (2x − 3) out. Keep the 2 in front.
2(2x − 3)(2x + 1) = 0
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Solve Step 10 of 10
Set each factor equal to zero The factor 2 can never be 0, so it gives no solution. Set the other two factors equal to 0.
2x − 3 = 0 or 2x + 1 = 0
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 24 · Done Done
Solution Nice work. Press Back to Questions to choose another question.
x = 3 2 x = −1 2
Original equation: 5x ² − 2x − 6 = −3x ² + 6x
Question 25 7 steps
Solve by factoring Try the first step on paper, then press Next to check it.
x ³ − 5x ² + 6x = 0
Question 25 · Standard form Step 1 of 7
Check standard form 0 is already alone on the right. This one starts with x ³, so look for a GCF that includes x .
x ³ − 5x ² + 6x = 0
Original equation: x ³ − 5x ² + 6x = 0
Question 25 · First, check for a GCF Step 2 of 7
First, check for a GCF Coefficients 1, −5 and 6 share only 1, but every term has x ; the lowest power is x x .
GCF(x ³, 5x ², 6x ) = x
Original equation: x ³ − 5x ² + 6x = 0
Question 25 · First, check for a GCF Step 3 of 7
Factor out the GCF Divide every term by x : x ³ ÷ x = x ², −5x ² ÷ x = −5x , 6x ÷ x = 6.
x x ² − 5x + 6) = 0
Original equation: x ³ − 5x ² + 6x = 0
Question 25 · Identify b and c Step 4 of 7
Identify b and c What is left inside the parentheses is a quadratic trinomial whose coefficient of x ² is 1 , so we can use the shortcut : find two numbers r and s whose product is c and whose sum is b . Then x ² + b x + c factors directly as (x + r )(x + s ) . Read b and c , keeping their signs.
a = 1, b = −5, c = 6
Original equation: x ³ − 5x ² + 6x = 0
Question 25 · Find the pair Step 5 of 7
Find the two numbers Find two numbers that multiply to 6 and add to −5 . The product is positive, so both numbers have the same sign . The sum is negative, so both numbers are negative. Factor pairs of 6: 1 · 6, 2 · 3. The pair −2 and −3 works.
(−2 )(−3 ) = 6 −2 + (−3 ) = −5
Original equation: x ³ − 5x ² + 6x = 0
Question 25 · Write the factors Step 6 of 7
Write the factors Put the two numbers, −2 and −3, straight into the pattern (x + ▢)(x + ▢). Keep the x in front. Adding a negative number is the same as subtracting.
x (x + (−2 ))(x + (−3 )) = 0 → x (x − 2)(x − 3) = 0
Original equation: x ³ − 5x ² + 6x = 0
Question 25 · Solve Step 7 of 7
Set each factor equal to zero Unlike a plain number such as 2 or 3, the factor x contains the variable
x = 0 or x − 2 = 0 or x − 3 = 0
Original equation: x ³ − 5x ² + 6x = 0
Question 25 · Done Done
Solution Nice work. Press Back to Questions to choose another question.
x = 0 or x = 2 or x = 3
Original equation: x ³ − 5x ² + 6x = 0
Word Problems Questions 26 – 28. Choose a question.
Jason jumped off of a cliff into the ocean in Acapulco while vacationing with some friends. His height as a function of time could be modeled by the function h (t ) = −16t ² + 16t + 480, where t is the time in seconds and h is the height in feet.
How long did it take for Jason to reach his maximum height? What was the highest point that Jason reached? Click a button to see the answer and the explanation.
Question 26 · a) Time to Maximum Height Word Problems
t = 0.5 seconds
The variable t stands for the input x and the variable h stands for the output y . Jason reaches the maximum height for t = −b /(2a ). a = −16, b = 16. t = −16 / (2(−16)) = 0.5 secondsb) Max height Question 26 · b) Highest Point Word Problems
h = 484 feet
The maximum height is reached at t = 0.5. Substitute 0.5 for t in the equation to find the maximum of h . h = −16(0.5)² + 16(0.5) + 480 = −4 + 8 + 480 = 484 feeta) Time to max You are trying to dunk a basketball. You need to jump 2.5 ft in the air to dunk the ball. The height that your feet are above the ground is given by the function h (t ) = −16t ² + 12t . What is the maximum height your feet will be above the ground? Will you be able to dunk the basketball?
Click a button to see the answer and the explanation.
Question 27 · Maximum Height Word Problems
h = 2.25 feet
The variable t stands for the input x and the variable h stands for the output y . The maximum height is the h -coordinate of the vertex. To find it we first need to know when it happens: the time t at the vertex. Only then can we substitute that time into h (t ). With a = −16 and b = 12, the maximum happens at t = −b /(2a ) = −12 / (2(−16)) = 0.375 seconds. Substitute 0.375 for t : h = −16(0.375)² + 12(0.375) = −2.25 + 4.5 = 2.25 feet Dunk? Question 27 · Can You Dunk? Word Problems
No, you cannot dunk
You need to jump 2.5 ft to dunk the ball. The maximum height of your feet is 2.25 ft. 2.25 < 2.5, so you will not be able to dunk the basketball. Max height If a toy rocket is launched vertically upward from ground level with an initial velocity of 128 feet per second, then its height h after t seconds is given by the equation h (t ) = −16t ² + 128t (if air resistance is neglected).
What will the height of the rocket be after 3 seconds? How long will it take the rocket to reach max height? What is the max height? Click a button to see the answer and the explanation.
Question 28 · a) Height After 3 Seconds Word Problems
h = 240 ft
The question is asking to find the value of h given the value of t . t = 3. Plug in 3 for t in the equation and simplify. h = −16(3)² + 128(3) = −144 + 384 = 240 ftb) Time to max c) Max height Question 28 · b) Time to Maximum Height Word Problems
t = 4 seconds
The variable t stands for the input x and the variable h stands for the output y . The rocket reaches the maximum height for t = −b /(2a ). a = −16, b = 128. t = −128 / (2(−16)) = 4 secondsa) Height at 3 s c) Max height Question 28 · c) Maximum Height Word Problems
h = 256 feet
The maximum height is reached at t = 4. Substitute 4 for t in the equation to find the maximum of h . h = −16(4)² + 128(4) = −256 + 512 = 256 feeta) Height at 3 s b) Time to max y = 2x ² − 4x − 2
Click a button to see the answer and the explanation.
Question 1 · Graph y = 2x ² − 4x − 2
Points: (1, −4), (0, −2), (2, −2), (3, 4), (−1, 4)
Start with the vertex (1, −4), the y -intercept (0, −2) and its reflection across x = 1: (2, −2). Choose x = 3: y = 2(3)² − 4(3) − 2 = 4 → (3, 4). Reflect across the axis: (−1, 4). Form Opening Width Vertex Axis of sym. y -int.Min/Max Dom & Range y = −1 2 x ² + 2x − 1
Click a button to see the answer and the explanation.
Question 2 · Graph y = −1 2 x ² + 2x − 1
Points: (2, 1), (0, −1), (4, −1), (−2, −7), (6, −7)
Start with the vertex (2, 1), the y -intercept (0, −1) and its reflection across x = 2: (4, −1). Choose x = −2: y = −1 2 Reflect across the axis: (6, −7). Form Opening Width Vertex Axis of sym. y -int.Min/Max Dom & Range y = −1 2 x + 2)² − 1
Click a button to see the answer and the explanation.
Question 3 · Graph y = −1 2 x + 2)² − 1
Points: (−2, −1), (0, −3), (−4, −3), (1, −11 2 11 2
Start with the vertex (−2, −1), the y -intercept (0, −3) and its reflection across x = −2: (−4, −3). Choose x = 1: y = −1 2 11 2 11 2 Reflect across the axis: (−5, −11 2 Form Opening Width Vertex Axis of sym. y -int.Min/Max Dom & Range y = (x − 1)² + 3
Click a button to see the answer and the explanation.
Question 4 · Graph y = (x − 1)² + 3
Points: (1, 3), (0, 4), (2, 4), (3, 7), (−1, 7)
Start with the vertex (1, 3), the y -intercept (0, 4) and its reflection across x = 1: (2, 4). Choose x = 3: y = (3 − 1)² + 3 = 7 → (3, 7). Reflect across the axis: (−1, 7). Form Opening Width Vertex Axis of sym. y -int.Min/Max Dom & Range 
